HDU4135(容斥原理)

Co-prime

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4090 Accepted Submission(s): 1619


[align=left]Problem Description[/align]
Given a number N, you are asked to count the number of integers between A and B inclusive which are relatively prime to N.
Two integers are said to be co-prime or relatively prime if they have no common positive divisors other than 1 or, equivalently, if their greatest common divisor is 1. The number 1 is relatively prime to every integer.

[align=left]Input[/align]
The first line on input contains T (0 < T <= 100) the number of test cases, each of the next T lines contains three integers A, B, N where (1 <= A <= B <= 1015) and (1 <=N <= 109).

[align=left]Output[/align]
For each test case, print the number of integers between A and B inclusive which are relatively prime to N. Follow the output format below.

[align=left]Sample Input[/align]

2
1 10 2
3 15 5

[align=left]Sample Output[/align]

Case #1: 5
Case #2: 10

思路：求1~m中与n互素的数的个数。

#include <cstdio>
#include <vector>
using namespace std;
typedef long long LL;
LL sieve(LL m,LL n)
{
vector<LL> divisor;
for(LL i=2;i*i<=n;i++)
{
if(n%i==0)
{
divisor.push_back(i);
while(n%i==0)    n/=i;
}
}
if(n>1)    divisor.push_back(n);
LL ans=0;
for(LL mark=1;mark<(1<<divisor.size());mark++)
{
LL odd=0;
LL mul=1;
for(LL i=0;i<divisor.size();i++)
{
if(mark&(1<<i))
{
mul*=divisor[i];
odd++;
}
}
LL cnt=m/mul;
if(odd&1)    ans+=cnt;
else ans-=cnt;
}
return m-ans;
}
LL a,b,n;
int main()
{
int T;
scanf("%d",&T);
for(int cas=1;cas<=T;cas++)
{
scanf("%lld%lld%lld",&a,&b,&n);
LL res=sieve(b,n)-sieve(a-1,n);
printf("Case #%d: ",cas);
printf("%lld\n",res);
}
return 0;
}